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Math Help - Maxima and Minima Trig Problem

  1. #1
    Member astartleddeer's Avatar
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    Maxima and Minima Trig Problem

    Hello,

    I'm having difficulty in understanding where the book has got these angles from:

    Question: Find the maximum and minimum values for  y = 3\sin\theta - 4\cos\theta\ in\ the\ range\ \theta\ to\ 2\pi

     \frac{d}{d\theta} (3\sin\theta - 4\cos\theta) = 4\sin\theta + 3\cos\theta

     4\sin\theta + 3\cos\theta = R\sin(\theta + \alpha)

     4\sin\theta + 3\cos\theta = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha

     R = \sqrt {4^2 + 3^2} = \pm5

     \alpha = \arctan \frac{3}{4} \approx 36.869^{\circ}

     \frac {dy}{d\theta} = 5\sin(\theta + 36.869^{\circ}) = \pm 5\ for\ a\ maximum\ or\ minimum\ value

     \theta \approx 90 - 36.869^{\circ} \approx 53.131^{\circ}

    Now, in my book the answer is 5 at  143^{\circ}8' for a maximum value, and -5 at  323^{\circ}8' for a minimum value. However, they have subtracted  \alpha from  180^{\circ} in order to obtain these pair of angles? I've drawn the function, and these values is where  5\sin(\theta + 36.869^{\circ}) = 0 . As a matter of fact, the maximum value first occurs at  53.131^{\circ} and the minimum value at  233.131^{\circ}.

    I've looked into this a bit deeper, and the only way a maximum or minimum value can occur at these computed angles - is if the first derivative is a negative cosine function.

    What have I done wrong?


    Thank you for your attention.
    Last edited by astartleddeer; July 20th 2012 at 08:38 AM.
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  2. #2
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    Re: Maxima and Minima Trig Problem

    first off, the problem states the interval of interest is 0 \le \theta < 2\pi ... you need to start thinking radians.

    y = 3\sin{\theta} - 4\cos{\theta}

    \frac{dy}{d\theta} = 3\cos{\theta} + 4\sin{\theta} = 0

    for \frac{dy}{d\theta} = 0 , \cos{\theta} and \sin{\theta} will have to have opposite signs ... which tells you \theta will have to be in quadrants II or IV.

    \tan{\theta} = -\frac{3}{4}

    \theta = \arctan\left(-\frac{3}{4}\right) + \pi, a quad II angle

    \theta = \arctan\left(-\frac{3}{4}\right) + 2\pi, a quad IV angle

    to finish, use the second derivative test for the two angles to determine max/min.
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    Re: Maxima and Minima Trig Problem

    for \frac{dy}{d\theta} = 0 , \cos{\theta} and \sin{\theta} will have to have opposite signs ... which tells you \theta will have to be in quadrants II or IV.

    \tan{\theta} = -\frac{3}{4}

    \theta = \arctan\left(-\frac{3}{4}\right) + \pi, a quad II angle

    \theta = \arctan\left(-\frac{3}{4}\right) + 2\pi, a quad IV angle

    to finish, use the second derivative test for the two angles to determine max/min.
    I don't understand. For  \frac {dy}{d\theta} why do we have to give sin and cos opposite signs to equal zero?

    If sin and cos are both made positive does this not make  \tan \theta = \frac{-3}{-4} = \frac {3}{4} ?

    Further, why have you added  \pi and  2\pi to  \tan = -\frac{3}{4} ?
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    Re: Maxima and Minima Trig Problem

    Quote Originally Posted by astartleddeer View Post
    I don't understand. For  \frac {dy}{d\theta} why do we have to give sin and cos opposite signs to equal zero?
    3(something) + 4(something else) = 0

    what does that tell you about the signs of (something) and (something else) ?

    If sin and cos are both made positive does this not make  \tan \theta = \frac{-3}{-4} = \frac {3}{4} ?
    if sine and cosine have the same sign, then tangent is positive ... the solutions for y' = 0 is where tangent is negative.

    Further, why have you added  \pi and  2\pi to  \tan = -\frac{3}{4} ?
    you need to research the range of the arctangent function.
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    Re: Maxima and Minima Trig Problem

    Quote Originally Posted by skeeter View Post
    3(something) + 4(something else) = 0

    what does that tell you about the signs of (something) and (something else) ?
    Then only one of them can be negative?

    If sine and cosine have the same sign, then tangent is positive ... the solutions for y' = 0 is where tangent is negative.
    Oops. I was meant to say, if sin and cos are both negative then the tangent is positive in the third quadrant?



    you need to research the range of the arctangent function.
    Oh my goodness! I had mistaken  \pi for  90^{\circ} ! Yes, I'm with you now. If the tangent is negative, it is  \pi + (-0.64^{c}) or  2\pi + (-0.64^{c}) . Second and fourth quadrant respectively.
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    Re: Maxima and Minima Trig Problem

    Ok let's do this.

     \frac{dy}{d\theta} = 4\sin\theta - 3\cos\theta = 0\ for\ a\ maximum\ or\ minimum\ value

     R = \sqrt{4^2 + (-3)^2} = \pm 5

     \theta = \arctan -\frac{3}{4} \approx -0.6435^{c}

     In\ the\ range\ 0 \leq \theta \leq 2\pi...\theta \approx \pi + (-0.6435^{c}) \approx 2.496^{c} and  2\pi + (-0.6435^{c}) \approx 5.639^{c}

     \frac{d^{2}y}{d\theta^2} = 3\sin\theta + 4\cos\theta

     \frac{d^{2}y}{d\theta^2} is a maximum for  2.496^{c}

     \frac{d^{2}y}{d\theta^2} is a minimum for  5.639^{c}

    Therefore, a minimum value occurs at  -5, 5.639^{c} and a maximum value occurs at  5, 2.496^{c}
    Last edited by astartleddeer; July 20th 2012 at 12:14 PM.
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    Re: Maxima and Minima Trig Problem

    Quote Originally Posted by astartleddeer View Post
    Ok let's do this.

    [tex] \frac{dy}{d\theta} = 4\sin\theta - 3\cos\theta = 0\ for\ a\ maximum\ or\ minimum\ value

     R = \sqrt{4^2 + (-3)^2} = \pm 5

     \theta = \arctan = -\frac{3}{4} \approx -0.6435^{\circ}

     In\ the\ range\ 0 \leq \theta \leq 2\pi...\theta \approx \pi + (-0.6435^{c}) \approx 2.496^{\circ} and  2\pi + (-0.6435^{c}) \approx 5.639^{c}

     \frac{d^{2}y}{d\theta^2} = 3\sin\theta + 4\cos\theta

     \frac{d^{2}y}{d\theta^2} is a maximum for  2.496^{c}

     \frac{d^{2}y}{d\theta^2} is a minimum for  5.639^{c}

    Therefore, a minimum value occurs at  -5, 5.639^{c} and a maximum value occurs at  5, 2.496^{c}

    Please confirm this.

    maximum function value of y = 5 at \theta = \arctan\left(-\frac{3}{4}\right) + \pi  \approx 2.498 RADIANS

    minimum function value of y = -5 at \theta = \arctan\left(-\frac{3}{4}\right) + 2\pi  \approx 5.640 RADIANS
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  8. #8
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    Re: Maxima and Minima Trig Problem

    I indicated radians with a small c for circular measure

    Ok, thank you for your time Skeeter. Done and dusted
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